b,\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)

=>\(\dfrac{ab+ac+bc}{abc}=0\)

=>ab+ac+bc=0

=>ab=-ac-bc

ac=-ab-bc

bc=-ab-ac

N=\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\)

N=\(\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)

N=\(\dfrac{1}{a^2-ab-ac+bc}+\dfrac{1}{b^2-ab-bc+ca}+\dfrac{1}{c^2-ac-bc+ab}\)

N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)

N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0

Bài 1:

a) Ta có:

\(\frac{-1}{3}< 0\)

\(\frac{1}{100}>0\)

\(\Rightarrow\frac{-1}{3}< \frac{1}{100}\)

b)Ta có;

 \(\frac{-231}{232}>-1\)

\(\frac{-1321}{1320}< -1\)

\(\Rightarrow\frac{-231}{232}>\frac{-1321}{1320}\)

c) Ta có:  

\(\frac{-27}{29}< 0\)

\(\frac{272727}{292929}>0\)

\(\Rightarrow\frac{-27}{29}< \frac{272727}{292929}\)

Bài 2:

\(a\left(b+1\right)=ab+a\)

\(b\left(a+1\right)=ab+b\)

Mà   \(a< b\)

\(\Rightarrow a\left(b+1\right)< b\left(a+1\right)\)

\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)

đơn giản, cứ áp dụng theo công thức là ra!!!!

những câu tích phân như này giải tay ko hề dễ, nên mình dùng table mò ra a=13,b=18,c=78 => a+b+c=109 :v